Key inertia - falling keys

[Mark Davidson]

RicB wrote:

> Hi folks. I wanted to try and turn this back to the origional question
> at hand.

...

> BW + FW  = ((SW x HR x WR) + WW) x KR

...

>
> Any force exerted upon the key to accelerate it and the rest the action
> then should be able to be viewed through this equation.  BW + FW become
> a <<ground zero>> as it were... a baseline, and ((SW x HR x WR) + WW) x
> KR define the mass and leverage (effective mass ?) being moved at any
> given time..... but the weight quanties would have to be translated to
> their respective moments of inertia, and the changing leverage through
> out the key stroke would have to be figured.

Hmm.  Weight translated to moment of inertia?  Good luck. Changing leverage?
Do you see significant changes in leverage throughout the key stroke?  Seems
to me changes would be small, maybe 2-3 percent at most.

>
> One way to approximate this, I had thought, was to see what sum of BW
> and FW it takes to balance the key at ten evenly spaced points through
> the key stroke. Obvioiusly you cant do this with UW and DW measurements,
> but there are ways of getting around that. Assuming then that you can
> find this  BW + FW  for these ten positions... you can interprete the
> resulting change either as changes in "effective weight" or as changes
> in leverage.... but not both.  If you interpret as changes in
> leverage... then you have  ten points you can plot on a graph which will
> show the leverage on the one axis against the position of the key on the
> other. Useing simple regression math you should be able to find an
> approximate equation for the leverage throughout the whole keystroke.
> Yes ??

Sounds possible (a stretch), but I think friction variation would be more
significant than any changes in leverage.

>
> Once you have that, and moments of inertia for each part it becomes
> easier to figure the amount of force needed to accellerate an action to
> any given velocity.

Well, you'd need the radii too, to figure all the torques, but OK.

<<FW>> seen as key inertia will gradualy reverse
> from a positive to a negative number as key velocity approaches and
> exceeds 9.8 m/sec^2. The others <<weights>> will keep the same sign.

Lost me here.  So OK.  There's this concept that if  the hole where the key
lead lives falls at rate of gravitational acceleration then the lead won't
exert force on the key.  As long as you ignore left to right acceleration
and rotation of the keylead, it might even be approximately true.  But
probably not useful.

But what is this changing sign business?  I guess you're saying that the key
is the only part that's pushing in the direction you want to go, and at some
point you start pushing it faster than the  it will fall.

Allllllrrrrrright.  Had some coffee.  Lightbulb went on. Let's do this
right.  Suppose a typical heavily leaded key has a moment of inertia of 25
kg cm^2,  FW of .035kg (35 grams), and radius from balance rail to front of
25 cm.  If we remove the top action, and just let the key fall, how fast
will it accelerate?  We use the torque(T) = moment of inertia(I) * angular
acceleration(A).  Also assume near horizontal so we don't have to do cosine
stuff.

Torque is .035kg * 980cm/sec/sec * 25 cm = 857 kg cm / sec / sec
A = T / I = 857 / 25 = 34.3 radians / sec / sec

If the radius of the key front is 25 cm, this means the key front falls at
25 * 34.3 cm/sec/sec = 857 cm/s/s.  A keylead halfway out falls at half that
rate.

So your concept is right, that at some rate of acceleration the key just
FALLS and no force is required to move it.  And only if you want to
accelerate it faster than that rate will the keyleads start being felt as
resistance.  The keyleads will be falling at much less than 9.8m/s/s though.
Bottom line is you only need to look at the key, not the rest of the action
to find this rate.

>
> Ok.. assuming you can do all this... it should be easier to compare
> assist spring actions with lead counterbalanced actions yes ?

One step closer.

-Mark